MC 302 – GRAPH THEORY – 12/3/13 –SOLUTIONS TO HW #5 – Due Friday, 11/22/13, by 3 PM
Individual or team of 2 – 34 POINTS TOTAL
1.
The picture at the right demonstrates that a soccer ball is essentially a convex
polyhedron with 12 pentagon faces and 20 hexagon faces. The dodecahedron
has 12 pentagon faces and no other faces. In this problem you’ll prove that any
convex polyhedron whose faces are either pentagons or hexagons must have
exactly 12 pentagon faces.
a. First a little geometry. We know that every vertex of a polyhedron must have degree at least 3,
because at least 3 faces must meet at every vertex. If we add up the degrees of the angles
formed by the edges meeting at a vertex of a convex polyhedron, the sum must be strictly less
than 3600, because otherwise all the faces would lie on the same plane. For example, in a cube,
the three angles at each vertex are all 900, and the sum is 2700. The formula for the angle at each
vertex of an -sided polygon is 180
. Use this to show that if the faces
of a convex
polyhedron are all pentagons or hexagons, then at most 3 faces meet at a vertex. Thus in such a
polyhedron exactly 3 faces meet at each vertex, and so every vertex has degree 3.
b. Now let S (for ‘solid,’ or maybe ‘soccer ball’) be a convex polyhedron, whose faces are all either
pentagons or hexagons. Let = the number of vertices, = the number of edges, = the
number of pentagon faces, and = the number of hexagon faces (thus = + is the total
number of faces). Use the Degree-Sum Formula to eliminate from Euler’s Formula, and derive
the equation + − = .
c. We know that there is also a Face-Sum formula: = ∑ ( ). Rewrite the summation of the
face degrees in terms of = number of faces with degree 5 and = number of faces with
degree 6. Then use the Face-Sum Formula to eliminate from Euler’s Formula, and derive the
equation − − = .
d. Use the two equations from parts (c) and (d) to prove = .
Note: If you substitute = 12 into either of the previous equations, you get = + ", or
equivalently, = − ". When ℎ = 0 and = 20, we’ve got the dodecahedron. When ℎ = 20 and
= 60, we’ve got the soccer ball. From surfing the web, I learned that any nonnegative value of ℎ,
except = , is possible (i.e., there is some polyhedron with that number of hexagons and 12
pentagons), and that there is connection between this and fullerenes in chemistry, where each vertex
represents a carbon atom. That’s the limit of my knowledge, but feel free to go and explore!
SOLUTION TO #1: (16 pts: 4 each part; correction sent 11/20: all polygons are regular)
a) By the given formula, the angle at each vertex of a pentagon is 1080, and the angle at each vertex of
a hexagon is 1200. Thus the minimum total angle sum we could have with any combination of four or
more pentagons and hexagons is at least 4(108o) = 432o> 360o. Hence we can’t have four or more
faces meet at a vertex if each face is a pentagon or hexagon. Note that three hexagon angles at a
vertex add up to exactly 3600, so that’s impossible. It is possible to have three pentagons at a vertex
(3240), two pentagons and a hexagon (3360), or one pentagon and two hexagons (3480).
b) By the Degree-Sum Formula, since every vertex has degree 3, 2 = ∑ deg(() = 3, so = 3/2.
Substituting this into Euler’s Formula gives 2 = − +
+ℎ =−
+
+
+ℎ =
+
+ ℎ.
Multiplying the first and last equal terms by 2 gives 4 = − + 2 + 2ℎ, or equivalently,
2 + 2ℎ − = 4.
c) When we add up the degrees of the faces, we count 5 for each pentagon and 6 for each hexagon, so
2 = ∑ deg(-. ) = 5 + 6ℎ. Thus =
2=−+
+ℎ =
01
+ 3ℎ. Plugging this into Euler’s Formula gives:
01
+1
− − 3ℎ + + ℎ = − − 2ℎ.
Multiplying the first and last equal terms by 2 gives 4 = 2 − 3 − 4ℎ, or equivalently,
2 − 3 − 4ℎ = 4.
d) If we multiply the equation from part (b) by 2 and add the resulting equation and the equation from
part (c), we get (4 + 4ℎ − 2) + (2 − 3 − 4ℎ) = 12. This simplifies to = 12.
2. Do the exercise that we discussed in class today. Display a subgraph of the 4-cube 2 that is a
subdivision of 34 or 3, . Show only the subgraph, with its vertices labeled with 4-tuples of zeroes and
ones. Draw your picture so that it’s visually obvious that the subgraph is a subdivision of 60 or 6+,+.
SOLUTION TO #2: (6 pts)
A subdivided 6+,+ subgraph is shown below left within 78 , and below right by itself. The red square vertices
and blue diamond vertices indicate the two partite sets of 6+,+, the small black circle vertices are the degree2 “helper” vertices, and the red-to-blue paths are shown in green, pink, and brown. The red vertices are
{0110, 1010, 1001} and the blue vertices are {0000, 1011, 1110}. The nine paths from red to blue are:
0110-0010-0000, 0110-1110, 0110-0111-0011-1011, 1010-1110, 1010-1011, 1010-1000-0000, 1001-1011, 1001-0001-0000,
and 1001-1101-1111-1110.
3. p. 187, 5.6.1 and 5.6.2 (these two exercises are related). Please draw each of the dual graphs 9:∗ and 9∗
twice, once superimposed on the original graphs and then a drawing of just the dual graph. To prove
that 9: and 9 are isomorphic, it suffices to label the vertices in each graph so that the labeled graphs
are identical.
SOLUTION TO #3: (12 pts: 6 pts--5.6.1, 6 pts—5.6.2; correction sent 11/16: use Fig. 5.37 for both problems)
Solution to 5.6.1: The original graphs of Fig. 5.37 are shown below to the left, with their dual graphs
drawn on top of them. To the right the dual graphs are displayed alone.
Solution to 5.6.2: The first picture below labels the vertices of the two original graphs so that the
resulting labeled graphs are identical, i.e., each vertex has the same neighbors in each graph; thus
the original graphs are isomorphic. It is clear that the duals (shown above right) are not isomorphic,
because they have different degree sequences: (4, 4, 3, 3, 3, 3) and (5, 3, 3, 3, 3, 3).